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Unformatted text preview: T T = 2 k L = 2(6 . 67 10 9 m 1 )(0 . 010)(750 10 12 m) = . 10 . There is a 10% decrease in the transmission coecient. (c) The change in the transmission coecient is given by T = dT dE E = 2 Le 2 kL dk dE E = 2 LT dk dE E . Now, dk/dE = dk/dU = k/ 2( U E ), so T T = kL E U E = (5 . 0) (0 . 010)(5 . 1 eV) 6 . 8 eV 5 . 1 eV = 0 . 15 . There is a 15% increase in the transmission coecient....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Mass

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